Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line. 
Show that $(1+\frac{x}{n})^n \rightarrow e^x$ uniformly on any bounded interval of the real line.

I am trying to argue from the definition of uniform convergence for a sequence of real-valued functions, but am struggling quite a lot. My efforts so far have concentrated on trying to find a sequences,
${a_n}$ which tends to zero, such that 
$$|(1+\frac{x}{n})^n -e^x |\leq a_n$$ for all $n$. But I have been unsuccessful thus far. All help is greatly appreciated.
 A: To find your sequence $(a_n)$ where $|(1 + x/n)^n - e^x| \leqslant a_n \to 0$ -- proving uniform convergence on any bounded interval -- use the inequality $\ln(1+y) \leqslant y$.
We have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} \leqslant  \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$
Take $y = x/n$. It follows that for $n$ sufficiently large
$$1 + \frac{x}{n} \leqslant e^{x/n} \leqslant \left(1 - \frac{x}{n}\right)^{-1},$$
and
$$\left(1 + \frac{x}{n}\right)^n \leqslant e^x \leqslant \left(1 - \frac{x}{n}\right)^{-n}.$$
The second inequality implies that
$$e^{-x} \geqslant \left(1 - \frac{x}{n}\right)^{n}.$$
Using Bernoulli's inequality $(1 - x^2/n^2)^n \geqslant 1 - x^2/n.$
Hence,
$$0 \leqslant e^{x} - \left(1+ \frac{x}{n}\right)^n = e^{x}\left[1 - e^{-x}\left(1+ \frac{x}{n}\right)^{n}\right]\\ \leqslant e^{x}\left[1 - \left(1+ \frac{x}{n}\right)^{n}\left(1- \frac{x}{n}\right)^{n}\right]\\= e^{x}\left[1 - \left(1- \frac{x^2}{n^2}\right)^{n}\right]\leqslant e^{x}\frac{x^2}{n}.$$
Therefore, for all $x \in [0,K]$, we have as $n \to \infty$  
$$0 \leqslant \left|e^{x} - \left(1+ \frac{x}{n}\right)^n\right| \leqslant e^K\frac{K^2}{n} \rightarrow 0.$$
An almost identical argument for $y \geqslant 0$ shows that
$$0 \leqslant e^{-y} - \left(1- \frac{y}{n}\right)^n \leqslant  e^{-y}\frac{y^2}{n}.$$
Thus if $-y = x \in [-L,0]$ we have
$$0 \leqslant \left|e^{x} - \left(1+ \frac{x}{n}\right)^n\right| = \left|e^{-y} - \left(1- \frac{y}{n}\right)^n\right| \leqslant \frac{L^2}{n} \rightarrow 0.$$
proving uniform convergence on any bounded interval.
A: 
Herein, we present an approach that for any given $\epsilon>0$, produces a number $N$, which depends on $\epsilon$ and not $x$, such that $\displaystyle \left|e^x-\left(1+\frac xn\right)^n\right|<\epsilon$ whenever $n>N$.


To do this we will use the inequalities, which I established in THIS ANSWER using only the limit definition of the exponential function and Bernoulli's Inequality.  The inequalities used in the ensuing analysis are


$$\bbox[5px,border:2px solid #C0A000]{e^x\le 1+x} \tag 1$$


for $x<-1$ and


$$\bbox[5px,border:2px solid #C0A000]{\log(x)\ge \frac{x-1}{x}} \tag 2$$


for $x>0$.


To this end, we proceed.


We assume that $x\in [a,b]$ and that $\epsilon>0$ is given. Furthermore, we will choose $n$ such that $n>-x$ for all $x\in[a,b]$.
Using $(1)$ and $(2)$ we can write
$$\begin{align}
\left|e^x-\left(1+\frac xn\right)^n\right|&=\left|e^x-e^{n\log\left(1+\frac xn\right)}\right|\\\\
&\le \left|e^x-e^{\frac{x}{1+x/n}}\right|\\\\
&=e^x\,\left|1-e^{-x^2/(x+n)}\right|\\\\
&\le e^x\,\left|\frac{x^2}{x+n}\right|\\\\
&\le e^{b}\frac{|\max^2(a,b)|}{n+a}\\\\
&<\epsilon
\end{align}$$
whenever $ \displaystyle n>\frac{e^b\,\max^2(a,b)}{\epsilon}-a$.  We take $\displaystyle N(\epsilon)=1+\left\lfloor \frac{e^b\,\max^2(a,b)}{\epsilon}-a \right\rfloor$ and we are done!
A: If you're not familiar with Dini's theorem (I wasn't), work from the power series of $\exp$.  Observe that
$$(1+x/n)^n = \sum_{k=0}^n {n \choose k} \frac{x^k}{n^k}$$
and that
$${n \choose k} \frac{1}{n^k} = \frac{n!}{n^k(n-k)!k!} \to \frac{1}{k!}$$
This means that each coefficient in the expansion of $g_n(x) = \exp(x) - \left(1+\frac{x}{n}\right)^n$ goes to $0$ as $n$ goes to infinity.  Also, all of the coefficients are positive and less than $\frac{1}{k!}$.  
Now, consider the interval $[-R,R]$.  Choose $k> 0$ such that $\sum_{s=k+1}^\infty \frac{R^s}{s!} < \frac{\epsilon}{2}$ (why can you do this?)  For $n$ sufficiently large, we can make each of the first $k+1$ coefficients smaller than $\frac{\epsilon}{2(k+1)(1+R^k)}$, so for any $x \in [-R,R]$,
$$|g_n(x)| \le  \sum_{s=0}^k \frac{\epsilon R^s}{2(k+1)(1+R^k)} + \sum_{s=k+1}^\infty \frac{R^s}{s!} \le \frac{\epsilon}{2} + \frac{\epsilon}{2}.$$
This same reasoning extends (with minor changes) to any bounded interval $[a,b]$ of $\mathbb{R}$ (and even half-open or open intervals).
